For the reaction between ammonia and hydrochloric acid to produce ammonium chloride; What volume of ammonia is used (at 20 Celsius and 760mmHg) if 12.5gmof ammonium chloride are produced in the reaction? How many molecules of ammonia does this represent?
NH3 + HCl ==> NH4Cl
mols NH4Cl = grams/molar mass = approx 12.5/53.5 = approx 0.23
Since 1 mol NH4Cl requires 1 mol NH3, plug 0.23 for mols into PV = nRT and solve for V in L.
To determine the volume of ammonia used in the reaction and the corresponding number of ammonia molecules, we need to follow a few steps:
Step 1: Write and balance the chemical equation:
NH3 + HCl → NH4Cl
Step 2: Convert the given mass of ammonium chloride to moles:
Given mass of NH4Cl = 12.5 g
To convert grams to moles, we need the molar mass of NH4Cl,
Molar mass of NH4Cl = (1 * 1) + (4 * 1) + (1 * 35.5) = 53.5 g/mol
Number of moles of NH4Cl = Given mass / Molar mass
Number of moles of NH4Cl = 12.5 g / 53.5 g/mol = 0.234 mol
Step 3: Use the stoichiometry of the balanced equation to determine the moles of ammonia used:
From the balanced equation, we see that 1 mole of NH3 reacts with 1 mole of NH4Cl.
Therefore, the number of moles of NH3 used = number of moles of NH4Cl
Number of moles of NH3 = 0.234 mol
Step 4: Convert moles of ammonia to volume using the ideal gas law:
The ideal gas law equation is:
PV = nRT
Given:
T = 20 °C = 293 K (convert to Kelvin)
P = 760 mmHg
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
Rearrange the ideal gas law equation to solve for volume:
V = nRT / P
V = (0.234 mol) * (0.0821 L·atm/(mol·K)) * (293 K) / (760 mmHg)
Step 5: Convert the volume from mmHg to liters:
1 atm = 760 mmHg
1 L = 1000 mL
V = (0.234 mol) * (0.0821 L·atm/(mol·K)) * (293 K) / (760 mmHg) * (1 atm / 760 mmHg) * (1000 mL / 1 L)
V ≈ 8.18 mL
Therefore, the volume of ammonia used in the reaction is approximately 8.18 mL.
To calculate the number of molecules of ammonia, we can use Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 molecules.
Number of molecules of NH3 = Number of moles of NH3 * Avogadro's number
Number of molecules of NH3 = 0.234 mol * (6.022 x 10^23 molecules/mol) ≈ 1.41 x 10^23 molecules
Therefore, the number of molecules of ammonia represented by 12.5 g of ammonium chloride is approximately 1.41 x 10^23 molecules.